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Gravitational force is a fundamental force of nature that draws two masses together. It is given by Newton's law of universal gravitation, which can be defined by the formula:

• \( F = \frac{{G M_1 M_2}}{{d^2}} \)

Here, \( F \) is the force of attraction between the masses, \( G \) is the gravitational constant, \( M_1 \) and \( M_2 \) are the masses, and \( d \) is the distance between them.
In our exercise, three stars each experience gravitational attractions from the other two. These forces work in tandem, pulling each star towards the center of their equilateral triangle.
This mutual attraction is key to achieving the necessary centripetal force for their circular motion, as it not only binds the stars together but causes them to rotate together symmetrically.

An equilateral triangle is a particularly symmetric shape where all sides have equal length and all angles are equal, each measuring 60 degrees. In the realm of physics problems, this symmetry simplifies calculations.

• Equilateral triangles are used to simplify analyses involving symmetry.
• They allow easy computation of certain properties due to predictable equal forces and angles.

In this problem, the concept is central, as it helps determine the distribution of the gravitational forces between the stars, simplifying the calculation for the net inward force.
Another important aspect from the equilateral triangle is its circumradius, which in this case is crucial to finding the effective radius of the stars' orbit. The circumradius \( r \) is given by:

• \( r = \frac{L}{\sqrt{3}} \)

This value is important because it helps determine the stars' circular path around the triangle's center.

Circular motion involves an object moving in a circle at a constant speed. It requires a net inward force, called the centripetal force, to maintain the object's circular path.

• Centripetal force keeps objects moving in a circle.
• It always points toward the center of the circle.

In our exercise, the centripetal force is provided by gravitational attraction between the stars. To calculate it:

• The force formula is \( F_c = \frac{M v^2}{r} \), where \( v \) is tangential speed and \( r \) is the radius to the rotational center.

For these stars, the radius \( r \) is determined by the equilateral triangle arrangement, \( \frac{L}{\sqrt{3}} \), and the gravitational pull gives the required force for their motion. This results in the motion described by our solution, finding each star's speed as \( v = \sqrt{\frac{3G M}{L}} \).
This speed is an outcome of the balance between gravitational attraction and the necessity of maintaining circular motion around the central point.